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91Ó°ÊÓ

A random sample of 100 faculty at a university gives the results shown below for professorial rank versus gender. a. Test for a relationship at the \(5 \%\) level using a chi-squared statistic. b. Test for a relationship at the \(5 \%\) level using logistic regression. c. Compare the \(P\)-values in parts (a) and (b). Is this in accord with your expectations? Explain. d. Interpret your results. Assuming that today's assistant professors are tomorrow's associate professors and professors, do you see implications for the future? $$ \begin{array}{lcc} \hline \text { Rank } & \text { Male } & \text { Female } \\ \hline \text { Professor } & 25 & 9 \\ \text { Assoc Prof } & 20 & 8 \\ \text { Asst Prof } & 18 & 20 \\ \hline \end{array} $$

Short Answer

Expert verified
Chi-square and logistic regression may show different p-values; it helps analyze the rank-gender relationship. Implications suggest potential changes in future gender distributions as ranks progress.

Step by step solution

01

Set Up the Hypotheses for Chi-Square Test

To test for a relationship at the 5% level using a chi-squared statistic, we set up the null hypothesis \( H_0 \): There is no association between professorial rank and gender. The alternative hypothesis \( H_a \): There is an association between professorial rank and gender.
02

Compute Expected Frequencies

For each cell in the contingency table, calculate the expected frequency as \( E = \frac{(row\ total \times column\ total)}{grand\ total} \). Fill in these values for each combination of rank and gender.
03

Compute Chi-Squared Statistic

For each cell, calculate the contribution to the chi-squared statistic using \( \chi^2 = \sum \frac{(O - E)^2}{E} \), where \( O \) is the observed frequency and \( E \) is the expected frequency. Sum these values to get the total chi-squared statistic.
04

Determine Degrees of Freedom and Critical Value

Calculate the degrees of freedom with the formula \((r - 1)(c - 1)\), where \( r \) is the number of rows and \( c \) is the number of columns. Use a chi-squared distribution table to find the critical value for a 5% significance level.
05

Compare Statistic to Critical Value

Compare the calculated chi-squared statistic to the critical value to decide whether to reject the null hypothesis. If the statistic is greater, reject \( H_0 \).
06

Set Up Logistic Regression

Logistic regression can be set up with gender as the dependent variable and professorial rank as the independent variable to test for a relationship. Use statistical software to compute the logistic regression model.
07

Obtain P-Value from Logistic Regression

Retrieve the p-value from the logistic regression output. This p-value will indicate if there's a significant relationship at the 5% level.
08

Compare P-Values from Both Tests

Compare the p-values obtained from the chi-squared test and the logistic regression. Discuss whether they support each other and align with expectations.
09

Interpret the Results and Future Implications

Based on the presence or absence of gender differences across ranks, discuss implications for future gender distributions in higher academic ranks, considering potential future promotions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chi-squared test
The Chi-squared test is a statistical method used to determine if there is a significant association between two categorical variables. Imagine a grid, or contingency table, where you count occurrences of different types—that's what the Chi-squared test works with.

Here's how it works in simple steps:
  • Step 1: Define your hypotheses. The null hypothesis (\(H_0\)) proposes no association between the variables—in this case, no link between professorial rank and gender.
  • Step 2: Calculate expected frequencies for each cell in the table using the formula \(E = \frac{(\text{row total} \times \text{column total})}{\text{grand total}}\). This tells us what we'd expect to see if \(H_0\) is true.
  • Step 3: Use the formula \(\chi^2 = \sum \frac{(O - E)^2}{E}\) for observed (\(O\)) and expected (\(E\)) frequencies to compute the Chi-squared statistic.
  • Step 4: Look up a chi-squared distribution table to compare your statistic against a critical value, considering the degrees of freedom \((\text{df} = (r - 1)(c - 1))\), to determine statistical significance.
If your Chi-squared statistic exceeds the critical value, you reject the null hypothesis and conclude a significant association exists.

This process offers insights into potential associations without specifying cause or effect.
Logistic regression
Logistic regression is a type of regression analysis used when the dependent variable is categorical. It predicts the probability of occurrence of an event by fitting data to a logistic curve.

Let's break this down:
  • Purpose: It helps determine relationships between a binary dependent variable and one or more independent variables. Here, the binary outcome is gender, related to professorial ranks.
  • Setup: Rank is the independent variable, and gender is the dependent variable. Logistic regression analyzes how rank might predict gender.
  • Output: The result includes coefficients, odds ratios, and a p-value, which you use to infer whether the independent variable predicts the dependent variable.
The logistic regression model provides interpretive clarity by indicating how likely one outcome is over another, given certain predictor conditions.

It is especially valuable because it adjusts for multiple covariates, offering a robust analysis in contrast to simpler methods.
Contingency table analysis
A contingency table, also known as a cross-tabulation, is a tool used in statistics to analyze the relationship between two or more categorical variables.

Here's a simple breakdown:
  • Structure: The table is organized into rows and columns, each representing categories from the variables being analyzed. In our example, rows show professorial ranks, and columns indicate gender.
  • Purpose: By counting each occurrence, it helps visualize potential associations or correlations between variables.
  • Analysis: Techniques like the Chi-squared test utilize contingency tables to assess whether observed differences between variables are statistically significant.
This method's strength lies in its simplistic visual representation, which eases comprehension when deciphering complex relationships.

It serves as a foundation for more complicated statistical tests, like Chi-squared, providing a preliminary assessment of data distributions.
P-value interpretation
The p-value is a cornerstone of hypothesis testing in statistics, indicating the probability that observed data could occur under the null hypothesis.

Here's how to understand it:
  • Definition: A p-value quantifies the evidence against a null hypothesis. A low p-value (<0.05, often considered statistically significant) suggests strong evidence to reject \(H_0\).
  • Comparison: In the Chi-squared test and logistic regression, a p-value helps assess the significance of observed relationships. Each method might produce different p-values, interpreting the association from varied angles.
  • Expectation: If both tests indicate low p-values, it strengthens confidence in the presence of a real association.
Interpreting the p-value helps determine statistical significance but not practical significance. Thus, even significant p-values should be considered in context with other evidence.

This interpretive tool is vital for making data-driven conclusions about the relationship studied.

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Most popular questions from this chapter

The article from which the data in Exercise 20 was obtained also gave the accompanying data on the composite mass/outer fabric mass ratio for highquality fabric specimens. \(\begin{array}{lllllll}1.15 & 1.40 & 1.34 & 1.29 & 1.36 & 1.26 & 1.22 \\ 1.40 & 1.29 & 1.41 & 1.32 & 1.34 & 1.26 & 1.36 \\ 1.36 & 1.30 & 1.28 & 1.45 & 1.29 & 1.28 & 1.38 \\ 1.55 & 1.46 & 1.32 & & & & \end{array}\) MINITAB gave \(r=.9852\) as the value of the Ryan- Joiner test statistic and reported that \(P\) value \(>.10\). Would you use the one-sample \(t\) test to test hypotheses about the value of the true average ratio? Why or why not?

The following data resulted from an experiment to study the effects of leaf removal on the ability of fruit of a certain type to mature ("Fruit Set, Herbivory, Fruit Reproduction, and the Fruiting Strategy of Catalpa speciosa," Ecology, 1980: 57-64). Does the data suggest that the chance of a fruit maturing is affected by the number of leaves removed? State and test the appropriate hypotheses at level \(.01\). $$ \begin{array}{lcc} \hline & \begin{array}{l} \text { Number } \\ \text { of Fruits } \\ \text { Matured } \end{array} & \begin{array}{c} \text { Number } \\ \text { of Fruits } \\ \text { Aborted } \end{array} \\ \hline \text { Control } & 141 & 206 \\ \text { Two leaves removed } & 28 & 69 \\ \text { Four leaves removed } & 25 & 73 \\ \text { Six leaves removed } & 24 & 78 \\ \text { Eight leaves removed } & 20 & 82 \end{array} $$

A statistics department at a large university maintains a tutoring center for students in its introductory service courses. The center has been staffed with the expectation that \(40 \%\) of its clients would be from the business statistics course, \(30 \%\) from engineering statistics, \(20 \%\) from the statistics course for social science students, and the other \(10 \%\) from the course for agriculture students. A random sample of \(n=120\) clients revealed \(52,38,21\), and 9 from the four courses. Does this data suggest that the percentages on which staffing was based are not correct? State and test the relevant hypotheses using \(\alpha=.05\).

a. Having obtained a random sample from a population, you wish to use a chi- squared test to decide whether the population distribution is standard normal. If you base the test on six class intervals having equal probability under \(H_{0}\), what should the class intervals be? b. If you wish to use a chi-squared test to test \(H_{0}\) : the population distribution is normal with \(\mu=.5, \sigma=.002\) and the test is to be based on six equiprobable (under \(H_{0}\) ) class intervals, what should these intervals be? c. Use the chi-squared test with the intervals of part (b) to decide, based on the following 45 bolt diameters, whether bolt diameter is a normally distributed variable with \(\mu=.5\) in., \(\sigma\) \(=.002 \mathrm{in}\). \(\begin{array}{llllll}.4974 & .4976 & .4991 & .5014 & .5008 & .4993 \\ .4994 & .5010 & .4997 & .4993 & .5013 & .5000 \\ .5017 & .4984 & .4967 & .5028 & .4975 & .5013 \\ .4972 & .5047 & .5069 & .4977 & .4961 & .4987 \\ .4990 & .4974 & .5008 & .5000 & .4967 & .4977 \\ .4992 & .5007 & .4975 & .4998 & .5000 & .5008 \\ .5021 & .4959 & .5015 & .5012 & .5056 & .4991 \\ .5006 & .4987 & .4968 & & & \end{array}\)

The results of an experiment to assess the effect of crude oil on fish parasites are described in the article "Effects of Crude Oils on the Gastrointestinal Parasites of Two Species of Marine Fish" (J. Wildlife Diseases, 1983: 253-258). Three treatments (corresponding to populations in the procedure described) were compared: (1) no contamination, (2) contamination by 1-year-old weathered oil, and (3) contamination by new oil. For each treatment condition, a sample of fish was taken, and then each fish was classified as either parasitized or not parasitized. Data compatible with that in the article is given. Does the data indicate that the three treatments differ with respect to the true proportion of parasitized and nonparasitized fish? Test using \(\alpha=.01\). $$ \begin{array}{lcc} \hline \text { Treatment } & \text { Parasitized } & \text { Nonparasitized } \\\ \hline \text { Control } & 30 & 3 \\ \text { Old oil } & 16 & 8 \\ \text { New oil } & 16 & 16 \end{array} $$

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